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Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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Updated on: 09 Apr 2012, 11:28
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Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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Originally posted by sacmanitin on 24 Oct 2009, 04:36.
Last edited by Bunuel on 09 Apr 2012, 11:28, edited 1 time in total.
Edited the question and added the OA




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Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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09 Apr 2012, 11:31




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Re: Is xy <6
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24 Oct 2009, 04:46
IMO B.
stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=8, y=3
stmt2: x is +ve, 8<y<8. Let us take max value of x =2/3 and max value of y=8, the product is <6. Now, let us take max value of x=2/3 and a negative value of y=8, the product is ve and hence < 6. One more try, take x=2/3 and y=5, the product is <6.



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Re: My mistake or OG's mistake ?
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06 Nov 2010, 20:50
Geronimo wrote: I don't really agree with OG's answer. What about you ? Is xy<6 ? (1) x<3 and y<2 (2) 1/2 < x < 2/3 and y^2<64 Is \(xy<6\)? (1) \(x<3\) and \(y<2\) > now, if both \(x\) and \(y\) are equal to zero then \(xy=0<6\) and the answer will be YES but if both \(x\) and \(y\) are small enough negative numbers, for example 10 and 10 then \(xy=100>6\) and the answer will be YES. Not sufficient. (2) \(\frac{1}{2}<x<\frac{2}{3}\) and \(y^2<64\), which is equivalent to \(8<y<8\) (not y<8, y>8. as written above) > even if we take the boundary values of \(x\) and \(y\) to maixmize their product we'll get: \(xy=\frac{2}{3}*8\approx{5.3}<6\), so the answer to the question "is \(xy<6\)" will always be YES. Sufficient. Answer: B.
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Re: Is xy < 6?
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28 Feb 2011, 01:56
Clearly 1) is sufficient. It says x is less than 3 and y is less than 2 so product can not be greater than 6.
Lets examine 2) x lies between 1/2 and 2/3 and y^2 is less than 64, or y lies within (8,8). On the extreme, xy can be just below 2/3*8 or ~5.33 so it will always be less than 6. Sufficient.
Answer should be D.



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Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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28 Feb 2011, 02:05



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Re: My mistake or OG's mistake ?
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01 Mar 2011, 23:13
How does y^2 <64 become y<8 and y>8?



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Re: My mistake or OG's mistake ?
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01 Mar 2011, 23:27
naveenhv wrote: How does y^2 <64 become y<8 and y>8? 1/2 < x < 2/3 and y^2<64 It is a general rule; \(y^2 < 64\) \(y < 8\) (Watch the less than (<) symbol) Means; 8<y<8 \(y^2 > 64\) \(y > 8\)(Watch the greater than (>) symbol) Means; y>8 or y<8 so; 8<y<8 1/2<x<2/3 Extreme values of xy 8*1/2 = 4 8*1/2 = 4 8*2/3 = 16/3 > 6 8*2/3 = 16/3 < 6 Thus; xy>6 xy<6 We found that; yes indeed xy<6. Sufficient.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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09 Apr 2012, 11:41
Bunuel, your explanations are so easy to understand. Thanks!
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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 <
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28 May 2013, 14:14
Bunuel wrote: mymbadreamz wrote: I didn't understand this one. could someone please explain why C is not the answer? thanks. Is \(xy<6\)? (1) \(x<3\) and \(y<2\) > now, if both \(x\) and \(y\) are equal to zero then \(xy=0<6\) and the answer will be YES but if both \(x\) and \(y\) are small enough negative numbers, for example 10 and 10 then \(xy=100>6\) and the answer will be NO. Not sufficient. (2) \(\frac{1}{2}<x<\frac{2}{3}\) and \(y^2<64\), which is equivalent to \(8<y<8\) > even if we take the boundary values of \(x\) and \(y\) to maixmize their product we'll get: \(xy=\frac{2}{3}*8\approx{5.3}<6\), so the answer to the question "is \(xy<6\)?" will always be YES. Sufficient. Answer: B. i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2? if a number times a number is less than 6......cant we just say use 1?



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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 <
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28 May 2013, 14:23
madzstar wrote: Bunuel wrote: mymbadreamz wrote: I didn't understand this one. could someone please explain why C is not the answer? thanks. Is \(xy<6\)? (1) \(x<3\) and \(y<2\) > now, if both \(x\) and \(y\) are equal to zero then \(xy=0<6\) and the answer will be YES but if both \(x\) and \(y\) are small enough negative numbers, for example 10 and 10 then \(xy=100>6\) and the answer will be NO. Not sufficient. (2) \(\frac{1}{2}<x<\frac{2}{3}\) and \(y^2<64\), which is equivalent to \(8<y<8\) > even if we take the boundary values of \(x\) and \(y\) to maixmize their product we'll get: \(xy=\frac{2}{3}*8\approx{5.3}<6\), so the answer to the question "is \(xy<6\)?" will always be YES. Sufficient. Answer: B. i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2? if a number times a number is less than 6......cant we just say use 1? On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another. Now, for x=y=0 we got an YES answer and for x=y=10 we got a NO answer, thus the statement is NOT sufficient. Of course we could use some other numbers to get an YES and a NO answers to prove that the statement is not sufficient: x=y=0 and x=y=10 are just examples of many possible sets. Hope it's clear.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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19 Oct 2016, 05:07
Geronimo wrote: Is xy < 6
(1) x < 3 and y < 2
(2) 1/2 < x < 2/3 and y^2 < 64 We need to determine whether the product of x and y is less than 6. Statement One Alone x < 3 and y < 2 Using the information in statement one we do not have enough information to determine whether the product of x and y is less than 6. For example, if x = 4 and y = 2, the product of x and y is 8, which is greater than 6. However, if x = 0 and y = 0, the product of x and y is 0, which is less than 6. We can eliminate answer choices A and D. Statement Two Alone1/2 < x < 2/3 and y^2 < 64 Using the information in statement two, we see that x is less than 2/3 and that y is less than 8. Thus, the maximum product of x and y is less than (2/3)(8) = 16/3, which is less 5.33 and thus less than 6. Since the maximum product of x and y is less than 6, statement two is sufficient to answer the question. Note that we did not even consider the case when 8 < y < 0 because in that case, xy will be negative and thus will be less than 6. Answer: B
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64
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